Do there exist distinct natural numbers $x, y, z, t$, all greater than or equal to $2$, such that $x \geq y + 2$, $z \geq t + 2$, and \[ \binom{x}{y} = \binom{z}{t}? \] (For natural numbers $n$ and $k$ with $n \geq k$, we define $\binom{n}{k} = \frac{n!}{k!(n-k)!}$.)